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Advanced Complex Analysis

Course Notes | Harvard University | Math 213a

C. McMullen

January 16, 2023

1 Basic complex analysis . . . . . . . . . . . . . . . . . . . . . . 2

2 The simply-connected Riemann surfaces . . . . . . . . . . . . 39

3 Entire and meromorphic functions . . . . . . . . . . . . . . . 59

4 Conformal mapping . . . . . . . . . . . . . . . . . . . . . . . 81

5 Elliptic functions and elliptic curves . . . . . . . . . . . . . . 109

Forward

Complex analysis is a nexus for many mathematical elds, including:

1. Algebra (theory of elds and equations);

2. Algebraic geometry and complex manifolds;

3. Geometry (Platonic solids;

at tori; hyperbolic manifolds of dimen- sions two and three);

4. Lie groups, discrete subgroups and homogeneous spaces (e.g.H=SL2(Z);

5. Dynamics (iterated rational maps);

6. Number theory and automorphic forms (elliptic functions, zeta func-

tions);

7. Theory of Riemann surfaces (Teichmuller theory, curves and their Ja-

cobians);

8. Several complex variables and complex manifolds;

9. Real analysis and PDE (harmonic functions, elliptic equations and

distributions). This course covers some basic material on both the geometric and analytic aspects of complex analysis in one variable. Prerequisites:Background in real analysis and basic dierential topology (such as covering spaces and dierential forms), and a rst course in complex analysis.

Exercises

(These exercises are review.)

1. LetTR3be the spherical triangle dened byx2+y2+z2= 1 and

x;y;z0. Let=z dxdz. (a) Find a smooth 1-formonR3such that=d. (b) Dene consistent orientations forTand@T. (c) Using your choices in (ii), computeR TandR @Tdirectly, and check that they agree. (Why should they agree?) 1

2. Letf(z) = (az+b)=(cz+d) be a Mobius transformation. Show the

number of rational mapsg:bC!bCsuch that g(g(g(g(g(z))))) =f(z) is 1, 5 or1. Explain how to determine which alternative holds for a givenf.

3. Let

Panznbe the Taylor series for tanh(z) atz= 0.

(a) What is the radius of convergence of this power series? (b) Show thata5= 2=15. (c) Give an explicit value ofNsuch that tanh(1) andPN

0anagree

to 1000 decimal places. Justify your answer.

4. Letf:U!Vbe a proper local homeomorphism between a pair of

open setsU;VC. Prove thatfis a covering map. (Hereproper means thatf1(K) is compact wheneverKVis compact.)

5. Letf:C!Cbe given by a polynomial of degree 2 or more. Let

1=ff(z) :f0(z) = 0g C

be the set of critical values off, letV0=f1(V1), and letUi=CVi fori= 0;1. Prove thatf:U0!U1is a covering map.

6. Give an example whereU0=U1is a normal (or Galois) covering, i.e.

wheref(1(U0)) is a normal subgroup of1(U1).

1 Basic complex analysis

We begin with an overview of basic facts about the complex plane and analytic functions. Some notation.The complex numbers will be denotedC. We let ;H andbCdenote the unit diskjzj<1, the upper half plane Im(z)>0, and the Riemann sphereC[ f1g. We writeS1(r) for the circlejzj=r, andS1for the unit circle, each oriented counter-clockwise. We also set = f0g andC=C f0g. 2

1.1 Algebraic and analytic functions

The complex numbers are formally dened as the eldC=R[i], where i

2=1. They are represented in the Euclidean plane byz= (x;y) =x+iy.

There are two square-roots of1 inC; the numberiis the one with positive imaginary part. An important role is played by the Galois involutionz7!z. We dene jzj2=N(z) =zz=x2+y2. (Compare the case of a real quadratic eld, whereN(a+bpd) =a2db2gives anindeniteform.) Compatibility ofjzj with the Euclidean metric justies the identication ofCandR2. We also see thatzis a eld: 1=z=z=jzj. It is also convenient to describe complex numbers by polar coordinates z= [r;] =r(cos+isin): Herer=jzjand= argz2R=2Z. (The multivaluedness of argzrequires care but is also the ultimate source of powerful results such as Cauchy's integral formula.) We then have [r1;1][r2;2] = [r1r2;1+2]: In particular, the linear mapsf(z) =az+b,a6= 0, ofCto itself, preserve angles and orientations. This formula should be provedgeometrically: in fact, it is a consequence of the formulajabj=jajjbjand properties of similar triangles. It can then be used to derive the addition formulas for sine and cosine (in Ah ors the reverse logic is applied). Algebraic closure.A critical feature of the complex numbers is that they arealgebraically closed; every polynomial has a root. (A proof will be reviewed below). Classically, the complex numbers were introducing in the course of solv- ingrealcubic equations. Staring withx3+ax+b= 0 one can make a Tschirnhaus transformation soa= 0. This is done by introducing a new variabley=cx2+dsuch thatPyi=Py2i= 0; even whenaandbare real, it may be necessary to chooseccomplex (the discriminant of the equation forcis 27b2+ 4a3.) It is negative when the cubic has only one real root; this can be checked by looking at the product of the values of the cubic at its max and min. Analytic functions.LetUbe an open set inCandf:U!Ca function.

We sayfisanalyticif

0(z) = limt!0f(z+t)f(z)t

3 exists for allz2U. It is crucial here thattapproaches zero through arbitrary values inC. Remarkably, this condition implies thatfis a smooth (C1) function. For example, polynomials are analytic, as are rational functions away from their poles. Note that anyreal linearfunction:C!Chas the form(v) =av+bv. The condition of analytic says thatDfz(v) =f0(z)v; in other words, thev part is absent. To make this point systematically, for a generalC1functionF:U!C we dene dFdz =12 dFdx +1i dFdy anddFdz =12 dFdx 1i dFdy

We then have

DF z(v) =dFdz v+DFdzv:

We can also write complex-valued 1-formdFas

dF=@F+@F=dFdz dz+dFdz dz ThusFis analytic i@F= 0; these are theCauchy-Riemannequations.

We note that (d=dz)z

n=nz n1; a polynomialp(z;z) behaves as if these variables are independent.

Sources of analytic functions.

1.Polynomials and rational functions.Using addition and multipli-

cation we obtain naturally the polynomial functionsf(z) =Pn

0anzn:

C!C. The ring of polynomialsC[z] is an integral domain and a unique factorization domain, sinceCis a eld. Indeed, sinceCis algebraically closed, fact every polynomial factors into linear terms.

It is useful to add the allowed value1to obtain the Riemann spherebC=C[f1g. Then rational functions (ratiosf(z) =p(z)=q(z) of rel-

atively prime polynomials, with the denominator not identically zero) determine rational mapsf:C!C. The rational functionsC(z) are the same as the eld of fractions for the domainC[z]. We setf(z) =1 ifq(z) = 0; these points are called thepolesoff.

2.Algebraic functions.Beyond the rational and polynomial functions,

the analytic functions includealgebraicfunctions such thatf(z) =pz

2+ 1. A general algebraic functionf(z) satisesP(f) =PN

0an(z)f(z)n=

0 for some rational functionsan(z); these arise, at least formally, when

4 one forms algebraic extension ofC(z). Such functions are generally multivalued, so we must choose a particular branch to obtain an ana- lytic function.

3.Dierential equations.Analytic functions also arise when one solves

dierential equations. Even equations with constant coecients, like y

00+y= 0, can give rise to transcendental functions such as sin(z),

cos(z) andez. Here are some useful facts about these familiar functions when extended toC: jexp(z)j= expRez cos(iz) = cosh(z) sin(iz) =isinh(z) cos(x+iy) = cos(x)cosh(y)isin(x)sinh(y) sin(x+iy) = sin(x)cosh(y) +icos(x)sinh(y): In particular, the apparentboundednessof sin(z) and cos(z) fails badly as we move away from the real axis, whilejezjis actually very small in the halfplane Rez0.

4.Integration.A special case of course is integration. WhileR(x2+ax+

b)1=2dxcan be given explicitly in terms of trigonometric functions, alreadyR(x3+ax+b)1=2dxleads one into elliptic functions; and higher degree polynomials lead one to hyperelliptic surfaces of higher genus. Note that the `periodicity' of the function in increases fromZ (trigonometric) toZ2(elliptic) toH1(g;Z) (hyperelliptic).

5.Power series.Analytic functions can be given concretely, locally, by

power series such asPanzn. Conversely, suitable coecients deter- mine analytic functions; for example,ez=Pzn=n!. For a more interesting example one can consider the partition function p(n) (satisfyingp(5) = 6 because 5 = 1+1+1+1+1 = 1+1+1+2 =

1 + 1 + 2 + 2 = 1 + 4 = 5), whose generating function satsies

1 X

0p(n)zn=1Y

1(1zn)1;

forz2. We remark thatQ(1 +an) converges ifPjanjconverges.

6.Riemann surfaces and automorphy.Another natural source of complex

analytic functions is functions that satisfy invariant properties such as 5 f(z+) =f(z) for all2, a lattice inC; orf(g(z)) =f(z) for all g2Aut(H). Theelliptic modular functionsf:H!Cwith have the property that f(z) =f(z+1) =f(1=z), and hencef((az+b)=(cz+d)) =f(z) for alla bc d2SL2(Z).

7.Geometric function theory.A complex analytic function can be

specied by a domainUC; we will see that for simply-connected domains (other thanCitself), there is an essentially unique analytic homeomorphismf: !U. WhenUtilesHorC, this is related to automorphic functions; and when@Uconsists of lines or circular arcs, one can also give a dierential equation forf.

8.Limits.We can also dene analytic functions by taking limits of poly-

nomials or other known functions. For example, consider the formula: e z= lim(1 +z=n)n: The triangle with vertices 0;1 and 1+i=nhas a hypotenuse of length

1 +O(1=n2) and an angle at 0 of+O(1=n2). Thus one ndsgeo-

metricallythatzn= (1 +i=n)nsatisesjznj !1 and argzn!; in other words, e i= cos+isin:

In particular,ei=1 (Euler).

1.2 Complex integration and power series

We now return to the general theory of analytic functions. LetUbe a compact,connected, smoothly bounded region inC, and letf:U!Cbe a continuous function such thatf:U!Cis analytic. We then have: Theorem 1.1 (Cauchy)For any analytic functionf:U!C, we haveR @Uf(z)dz= 0. Remark.It is critical to know thedenitionof such a path integral. (For example,f(z) = 1 is analytic, its average over the circle is 1, and yetR S

11dz= 0; why is this?)

If : [a;b]!Cparameterizes an arc, then we dene Z f(z)dz=Z b a f( (t))

0(t)dt:

6 Alternatively, we choose a sequence of pointsz1;:::;znclose together along , and then dene Z f(z) = limn1X

1f(zi)(zi+1zi):

(If the loop is closed, we choosezn=z1). This should not be confused with the integral with respect to arclength: Z f(z)jdzj= limXf(zi)jzi+1zij: Note that the formerdependson a choice of orientation of , while the latter does not. Proof of Cauchy's formula:(i) observe thatd(f dz) = (@f)dz dz= 0 and apply Stokes' theorem. (ii | Goursat). Cut the regionUinto small squares, observe that on these squaresf(z)az+b, and use the fact thatR @U(az+b)dz= 0.Aside: distributions.The rst proof implicitly assumesfisC1, while the second does not. (To see whereC1is used, suppose=udx+v dy andd= 0 on a squareS. In the proof thatR @S= 0, we integratevdy over the vertical sides ofSand observe that this is the same as integrating dv=dxdxdyover the square. But ifis notC1, we don't know thatdv=dx is integrable.) More generally, we say a distribution (e.g. anL1functionf) is (weakly) analytic ifRf@= 0 for every2C1c(U). By convolution with a smooth function (amollier), any weakly analytic function is a limit ofC1analytic functions. We will see below that uniform limits ofC1analytic functions areC1, so even weakly analytic functions are actually smooth. An important and standard computation in dierential forms shows that, forf2C10(C), we have Z C (@f)^dzz Z

C(epsilon)d(f(z)dz=z) =Z

1()f(z)z

dz 2if(0); and hence@(dz=z) = 2i0, as a current. More on the Cauchy{Riemann equations and with minimal smoothness assumptions can be found in [GM]. Cauchy's integral formula: Dierentiability and power series.Because of Cauchy's theorem, only one integral has to be explicitly evaluated in 7 complex analysis (hence the forgetability of the denition of the integral).

Namely, setting

(t) =eitwe nd, for anyr >0,Z S

1(r)1z

dz= 2i: By integrating between@Uand a small loop aroundp2U, we then obtain

Cauchy's integral formula:

Theorem 1.2For anyp2Uwe have

f(p) =12iZ @Uf(z)dzzp Now the integrand depends onponly through the rational function

1=(zp), which is innitely dierentiable. (It is theconvolutionoffj@U

and 1=z.) So we conclude thatf(p) itself is innitely dierentiable, indeed, it is approximated by a sum of rational functions with poles on@U. In particular, dierentiating under the integral, we obtain: f (k)(p)k!=12iZ @Uf(z)dz(zp)k+1(1.1) Ifd(p;@U) =R, the length of@UisLand sup@Ujfj=M, then this gives the bound: jakj= f (k)(p)k!

ML2Rk+1

In particular,

Pak(zp)khas radius of convergence at leastR, since limsupjakj1=k<1=R. This suggests thatfis represented by its power series, and indeed this is the case: Theorem 1.3Iffis analytic onB(p;R), thenf(z) =Pak(zp)kon this ball. Proof.We can reduce to the casez2B(0;1). Then forw2S1and xed zwithjzj<1, we have

1wz=1w

1 + (z=w) + (z=w)2+;

converging uniformly on the circlejwj= 1. We then have: f(z) =12iZ S

1f(w)dwwz=Xzk12iZ

1f(w)dww

k+1=Xa kzk as desired.8 Corollary 1.4An analytic function has at least one singularity on its circle of convergence. That is, iffcan be extended analytically fromB(p;R) toB(p;R0), then the radius of convergence is at leastR0. So there must be some obstruction to making such an extension, if 1=R= limsupjakj1=k. Example: Fibonacci numbers.Letf(z) =Panzn, whereanis thenth Fibonacci number. We have (a0;a1;a2;a3;:::) = (1;1;2;3;5;8;:::). Since a n=an1+an2, except forn= 0, we get f(z) = (z+z2)f(z) + 1 and sof(z) = 1=(1zz2). This has a singularity atz= 1= and thus limsupjanj1=n= , where = (1 +p5)=2 = 1:618:::is the golden ratio (slightly more than the number of kilometers in a mile).

Note thatP(an

n)zn=f(z)=(1 z) has radius of convergence j j>1, if we choose the constantto cancel the pole off(z) atz= 1=

Thusjan

nj !0. (In fact= =p5.) Theorem 1.5A power series represents a rational function i its coe- cients satisfy a recurrence relation. Aside: Pisot numbers.The golden ratio is an example of a Pisot number; as we have just seen, it has the property thatd( n;Z)!0 asn! 1. It is an unsolved problem to show that if >1 satisesd(n;Z)!0, thenis an algebraic number. Kronecker's theoremasserts thatPaiziis a rational function i deter- minants of the matricesai;i+j, 0i;jnare zero for allnsuciently large [Sa,xI.3] Question: why are 10:09 and 8:18 such pleasant times? [Mon]. How to compute.The power series for the arctangent is easy to evaluate by relating it toRdx=(1 +x2). Thus we get: f(z) = tan1(z) =zz3=3 +z5=5 +z7=7 + which suggest correctly that 4 = 113 +15 17 However the convergence is very slow, since the error afternterms is on the order of 1=n. 9 In general, to accelerate the convergence of the sumssn=sn(0) =Pn 0ai, we can setsn(1) = (sn+sn+1)=2 and take its limit instead. This will be especially ecacious for an alternating series. But why not repeat the process again and again? To this end we dene inductively: s n(k+ 1) =sn(k) +sn+1(k)2 If we have the terms (a0;:::;an) at our disposal, we can then compute e n=s0(n) = 2nnX 0 n k s n:

Forf(z) =Pakzkas above, we nd that

4 100X
0a k= 3:121:::but 4100X 02

100100

k a k= 3:141592653589793273:::; which agrees withto 16 decimal places (!) The reason this works is that if we writeF(y) =f(y=(1y)) =Pbkyk, thenen=Pn

0bk(1=2)k. Assumingfis analytic in the unit disk,Fis analytic

in the halfplane Re(y)1=2. But iff(z) is analytic in a neighborhood of z= 1, thenF(y) is analytic inB(0;r) for somer >1=2, and hence its power series converges geometrically fast aty= 1=2. In the case at hand,f(z) has singularities atz=i(and nowhere else). ThenF(z) has singularities ati=(1 +i), so its radius of convergence is

1=p2 which is bigger than 1=2. In fact we expect the error to be roughly of

size 2 n=2after takingnterms, and 250= 91016, consistent with the results above.

To justify this explanation, we must show that

e n=nX 0b k2k:(1.2)

To this end, rst note that:

1(1y)n+1=1X

0 n+k k y k: This can be seen by either dierentiating both sides, starting withn= 0, or by observing that the coecient ofykis the number of ways of writing 10 k=m1+mnwithmi0. Plugging intof(y=(1y)), we nd that b

0=a0and

b n+1=nX 0 n k a k+1: We can now show (1.2) by induction. We havee0=b0=a0, and e n+1en=s1(n)s0(n)2 = 2(n+1)nX 0 n kquotesdbs_dbs6.pdfusesText_11

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